Question

A wave travelling along the positive $x-$direction having maximum displacement along $y-$direction as $1\text{m}$, wavelength $2\pi \text{m}$ and frequency of $1/\pi \text{Hz}$ is represented by:

• A. $y=\sin (x-2t)$
• B. $y=\sin (2\pi x-2\pi t)$
• C. $y=\sin (10\pi x-20\pi t)$
• D. $y=\sin (2\pi x+2\pi t)$

Answer 1

The correct option is A $y=\sin (x-2t)$

Maximum vertical( $y-$direction) displacement of particle on string is the amplitude of wave
$\Rightarrow A=1\text{m}$
Wavelength$\left(\lambda \right)=2\pi \text{m}$,
frequency $\left(f\right)=\frac{1}{\pi }\text{Hz}$
Let the equation of wave be,
$y=A\sin (kx-\omega t+\varphi )$
where $k=\frac{2\pi }{\lambda }$ and $\omega =2\pi f$
$\Rightarrow y=1\times \sin (\frac{2\pi }{2\pi }x-\frac{2\pi }{\pi }t+\varphi )$
$\Rightarrow y=\sin (x-2t+\varphi )$
Now if at time $t=0$, particle on string at $x=0$ starts motion from its mean position i.e $y=0$ then phase constant $\varphi =0$.
Hence the wave equation will reduce to,
$y=\sin (x-2t)$
$\therefore$Option $\left(a\right)$ is correct.